Question: $g(x) = 4x^{3}-4x^{2}+3x+2(f(x))$ $h(t) = -7t^{3}+3t^{2}-4(f(t))$ $f(n) = 4n^{2}$ $ f(h(-2)) = {?} $
First, let's solve for the value of the inner function, $h(-2)$ . Then we'll know what to plug into the outer function. $h(-2) = -7(-2)^{3}+3(-2)^{2}-4(f(-2))$ To solve for the value of $h$ , we need to solve for the value of $f(-2)$ $f(-2) = 4(-2)^{2}$ $f(-2) = 16$ That means $h(-2) = -7(-2)^{3}+3(-2)^{2}+(-4)(16)$ $h(-2) = 4$ Now we know that $h(-2) = 4$ . Let's solve for $f(h(-2))$ , which is $f(4)$ $f(4) = 4(4^{2})$ $f(4) = 64$